Can someone explain to me the range for this particular function please
$$ y= \frac {2} {\sqrt{4-x^2}} $$
I'm in Year 10 so can someone explain this to me in a simple way.
Don't really get why it is y larger than or equal to 1
But I know the bottom cannot be equal to 0
For this particular function, lets consider the bottom first
We know that for the square root, $$ \sqrt {4-x^2} $$ that the values that this can exist only range from 0 to 2. This is because the value for x by itself can only exist up to x=2, because any larger would mean 4 subtracting number larger than 4, where the square root of a negative number DOES NOT EXIST. Hence the max limit of x is 2. This is mirrored on the negative x axis since x has been squared.
Therefore for, $$ \sqrt {4-x^2} $$, then $$ 0 \leq \sqrt {4-x^2} \leq 2$$
Then if you place this over the function in the question, we can sub the limits of this range to test for particular values, e.g. $$ y= \frac 2 2 $$ is equal to 1, and also $$ \frac 2 {\text{very small number close to zero}} $$ becomes a very large number headed towards infinity.
OBVIOUSLY, the 0 cannot be the denominator because this is invalid
As such the range of values exist from 1 to a very large number so, $$ y \geq 1 $$
If $x=0$ then $y=1$. For all other $x$ the denominator is smaller than $2$ so $y$ is greater than $1$
$2\ge \sqrt{4-x^2}$ for any $x\in(-2,2)$
indeed $4\ge 4-x^2$ means $x^2\ge 0$ which is always true in $(-2,2)$
The domain is $\{x \in \mathbb{R} \mid -2 < x < 2\}$.
Since the function has $y$-axis symmetry, to find the range, we only need to consider the restricted domain $\{x \in \mathbb{R} \mid 0 \le x < 2\}$.
As $x$ continuously increases from $0$ to $2$, not including $2$,
It follows that the range is $\{y \in \mathbb{R} \mid y \ge 1\}$.
